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Radius of the circle $ \vec{r}^2 + \vec{r} \cdot (2\hat{i} - 2\hat{j} - 4\hat{k}) - 19 = 0, \vec{r} \cdot (\hat{i} - 2\hat{j} + 2\hat{k}) + 8 = 0 $ is

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Radius of the circle $\vecr+\vecr·(2\hati-2\hatj-4\hatk)-19=0, \vecr·(\hati-2\hatj+2\hatk)+8=0$ is

Updated On: Apr 15, 2026

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The Correct Option is B

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Radius of the circle \( \vec{r}^2 + \vec{r} \cdot (2\hat{i} - 2\hat{j} - 4\hat{k}) - 19 = 0, \vec{r} \cdot (\hat{i} - 2\hat{j} + 2\hat{k}) + 8 = 0 \) is

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The Correct Option is B

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